Let $X$ and $Y$ be two smooth projective surfaces and $f:X\rightarrow Y$ a birational morphism. Let $E$ be the exceptional curve on $X$ that is contracted. Now EGA III Theorem 4.1.5 states that the functor $\mathcal{F}\mapsto f^*\mathcal{F}$ is an equivalence between categories of locally free $\mathcal{O}_Y$-modules of rank $r$ and locally free $\mathcal{O}_X$-modules $\mathcal{E}$ of the same rank such that the restriction of $\mathcal{E}$ to the formal completition $\mathcal{X}=X_{/E}$ is trival. Is it possible to generalize this Theorem for arbitrary smooth projective schemes over a field of characteristic zero with birational proper morphism between them?

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    $\begingroup$ Something about the formulation certainly must be changed, because the center $Z\subset Y$ can have nontrivial locally free sheaves. If $\mathcal{F}$ is a locally free sheaf on $Y$ whose restriction to $Y_{/Z}$ is nontrivial, then certainly the restriction of $f^*\mathcal{F}$ to $X_{/E}$ will also be nontrivial. If there is a result of the type you want, I suspect it will be found in Michael Artin's, "Algebraization of Formal Moduli, II". $\endgroup$ Jan 31, 2014 at 13:57
  • $\begingroup$ Also notice that every birational morphism in your setting is a blowing-up along some (possibly non-reduced ...) closed subscheme (see for instance Liu's book, chap. 8, Th. 1.24). $\endgroup$ Jan 31, 2014 at 14:02
  • $\begingroup$ @Aleksa: Unfortunately, I am too stupid to see how you conclude the equivalence of the above categories from the Formal Function Theorem (4.1.5). Could you please say a few words about this? $\endgroup$
    – boxdot
    Feb 11, 2014 at 18:37
  • $\begingroup$ @finite: Existence of vector bundles and global resolutions for singular surfaces. Compositio Math. 140 (2004), 717-728. Proposition 1.2 $\endgroup$
    – user45766
    Feb 13, 2014 at 14:47

1 Answer 1


If $f:X \to Y$ is a blowup with smooth center $Z$ and exceptional divisor $E$ then the subcategory $Lf^*(D(Y)) \subset D(X)$ of the derived category of coherent sheaves on $X$ identifies with the subcategory of objects such that their (derived) restriction to $E$ is trivial on fibers of $E$ over $Z$ (in the derived sense). For vector bundles you can forget about derived stuff --- a vector bundle on $X$ is a pullback from $Y$ if and only if it restricts trivially onto any fiber of $E$ over $Z$. Note that you don't need to consider the formal completion.

Further, if the morphism $f:X \to Y$ is a composition of blowups with smooth centers (if the strong factorization holds) then you can iterate the above description. If there is no strong factorization, probably you can use weak factorization instead (which always exists!) and also get some description.

EDIT. Here is a sketch of a proof of the statement about the criterion for a bundle to be pullback. Assume that $F$ is a vector bundle on $X$ and $F_{|E}$ restricts trivially to any fiber of $E = P_Z(N)$ over $Z$. Let $p:P_Z(N) \to Z$ be the projection and $O_E(-1)$ --- relative $O(-1)$. For any object $G \in D(Z)$ one has $$ Ext^\bullet(F,i_*(p^*G\otimes O_E(-k))) = Ext^\bullet(F_{|E},p^*G\otimes O_E(-k)) = H^\bullet(Z,Rp_*(F^\vee_{|E}\otimes p^*G(-k))) = 0 $$ for $k = 1,\dots,r(N)-1$ since $Rp_*(F^\vee_{|E}\otimes p^*G(-k)) = G\otimes^L Rp_*(F_{|E}(-k)) = 0$. Thus $F$ is contained in the first component of a semiorthogonal decomposition $$ D(X) = \langle i_*(O_E(1-r(N))\otimes p^*D(Z)), \dots, i_*(O_E(-1)\otimes p^*D(Z)), Lf^*(D(Y)) \rangle, $$ where $f:X \to Y$ is the blowup and $i:E \to X$ is the embedding. Thus $F \cong Lf^*F'$ for some object $F' \in D(Y)$. One can also check that $F'$ is a vector bundle.

  • $\begingroup$ Dear Sasha, could you please give an argument or a reference for your statement about the pullback of vector bundles from $Y$? $\endgroup$
    – boxdot
    Feb 7, 2014 at 14:15
  • $\begingroup$ I guess you can conclude this from Proposition 3.1 in 'Semiorthogonal decompositions for algebraic varieties' of Bondal and Orlov. $\endgroup$
    – Aleksa
    Feb 7, 2014 at 14:27
  • $\begingroup$ @finite: I added an argument. Aleksa is right, it is based on Orlov's blowup formula. $\endgroup$
    – Sasha
    Feb 7, 2014 at 19:15

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